A speeding car traveling at a constant velocity of 24 m/s passes a police care initially at rest beside the roadway.If the police car with constant acceleration of 3.0 m/s² immediately pursues the speeding car,how far down the road does it draw even with the speeding car? A)190 m B)290 m C)380 m D)770 m

Question

Quizplus · Accepted Answer

We can use the equations of motion to solve the problem. Let's assume that both cars meet after t seconds and they have covered a distance of d.

For the speeding car:
Initial velocity (u) = 24 m/s
Time (t) = t
Acceleration (a) = 0 m/s^2 (since the car is moving at a constant velocity)
Distance (d) = ut = 24t

For the police car:
Initial velocity (u) = 0 m/s
Time (t) = t
Acceleration (a) = 3.0 m/s^2
Distance (d) = ut + (1/2)at^2 = (3/2)t^2

Now, we need to find the value of t when both cars meet. We can equate the distance covered by both cars and solve for t.

24t = (3/2)t^2

t^2 - 16t = 0

t(t-16) = 0

t = 0 s (not possible) or t = 16 s

Therefore, both cars meet after 16 seconds.

We can now find the distance d covered by both cars using either of the above equations. Let's use the equation for the police car.

d = ut + (1/2)at^2 = (0)(16) + (1/2)(3)(16)^2 = 380 m

Therefore, the police car draws even with the speeding car after covering a distance of 380 m down the road.

The correct answer is C) 380 m.

A Speeding Car Traveling at a Constant Velocity of 24